Question: $h(t) = -6t-5(f(t))$ $f(t) = 3t$ $g(t) = -3t+f(t)$ $ f(g(-1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(-1)$ . Then we'll know what to plug into the outer function. $g(-1) = (-3)(-1)+f(-1)$ To solve for the value of $g$ , we need to solve for the value of $f(-1)$ $f(-1) = (3)(-1)$ $f(-1) = -3$ That means $g(-1) = (-3)(-1)-3$ $g(-1) = 0$ Now we know that $g(-1) = 0$ . Let's solve for $f(g(-1))$ , which is $f(0)$ $f(0) = (3)(0)$ $f(0) = 0$